Maximum Speed for Circular Turns: Radius Doubling Question Explained

[micjk]

Homework Statement

The maximum speed with which a car can take a circular turn of radius R is v. The maximum speed with which the same car, under the same conditions, can take a circular turn of radius 2R is

A. 2v
B. v√2
C. 4v
D. 2v√2

Homework Equations

v = (2πr)/T

The Attempt at a Solution

Since radius is directly proportional to the velocity, wouldn't the answer just be A. 2v because doubling the radius would double the velocity? The correct answer is B. and I have no idea as to why. Could someone please explain why. Thank you

 

[Mitchel Haas]

Homework Statement

The maximum speed with which a car can take a circular turn of radius R is v. The maximum speed with which the same car, under the same conditions, can take a circular turn of radius 2R is

A. 2v
B. v√2
C. 4v
D. 2v√2

Homework Equations

v = (2πr)/T

The Attempt at a Solution

Since radius is directly proportional to the velocity, wouldn't the answer just be A. 2v because doubling the radius would double the velocity? The correct answer is B. and I have no idea as to why. Could someone please explain why. Thank you

The maximum uniform circular velocity, is limited by the static frictional force of the tires on the pavement.
So, Newton's 2nd law gives us $$∑F = f = m a_{\text{rad}}=\frac{m v^2}{R}$$
The maximum friction for available is f_max = u_sm g.
So, you need to consider the radial acceleration, $$a_{\text{rad}}=\frac{v^2}{R}$$ Try again with taking these equations into account, keeping in mind that the coefficient of static friction, hence f, will be the same regardless of the speed.

 

[micjk]

The maximum uniform circular velocity, is limited by the static frictional force of the tires on the pavement.
So, Newton's 2nd law gives us $$∑F = m a_{\text{rad}}=\frac{m v^2}{R}$$
So, you need to consider the radial acceleration, $$a_{\text{rad}}=\frac{v^2}{R}$$ Try again with taking this equation into account, keeping in mind that the coefficient of static friction, hence f, will be the same regardless of the speed.

Ok I think I got it. Since in this case, the centripetal force is equal to the frictional force between the tires and the pavement, I can say that F_centripetal = F_frictional and therefore (m*v²)/2R = μ*m*a. I can solve for v and get v = √(2R * μ * a). I can also disregard μ since the frictional force will be constant. Then I can replace a with (v²)/R to get v = √(2R * v²/R. Simplifying that expression with give me maximum v = v√2
Is that correct?

 

[Mitchel Haas]

Ok I think I got it. Since in this case, the centripetal force is equal to the frictional force between the tires and the pavement, I can say that F_centripetal = F_frictional and therefore (m*v²)/2R = μ*m*a. I can solve for v and get v = √(2R * μ * a). I can also disregard μ since the frictional force will be constant. Then I can replace a with (v²)/R to get v = √(2R * v²/R. Simplifying that expression with give me maximum v = v√2
Is that correct?

Yes, this is correct, and you're using the proper reasoning. Be sure to distinguish between v₁ and v₂

 

[micjk]

Yes, this is correct, and you're using the proper reasoning. Be sure to distinguish between v₁ and v₂

Thank you!